dy/dx= dy/du*du/dx
A y=u^3/2 +1
dy/du=3/2 sqrtu
du/dx=4x
dy/dx=3/2 sqrt(u)*4x then
dy/dx=3/2 sqrt(2x^2 -2/3)*4x
now on the second,
dy/dx=dy/du*du/dx=(dy/du)/(dx/du)
That is two messy to try to type in detail. Find dy/du,dx/du and put it in the fraction. Algebra to simplify might be a challenge.
Find the dy/dx
A. y = u sqrt u + 1;
u = 2x^2 - 2/3
B. x = u/ (1 + u^3);
y = u^2 /(1 + u^3)
Can someone help me
2 answers
x = u/(1 + u^3)
y = u^2/(1 + u^3)
dy/dx = dy/du / dx/du
dy/du = u(2-u^3)/(1+u^3)^2
dx/du = (1-2u^3)/(1+u^3)^2
dy/dx = u(2-u^3)/(1-2u^3)
y = u^2/(1 + u^3)
dy/dx = dy/du / dx/du
dy/du = u(2-u^3)/(1+u^3)^2
dx/du = (1-2u^3)/(1+u^3)^2
dy/dx = u(2-u^3)/(1-2u^3)
7 answers